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Grade 12Thermal Physics

A cup of tea cool from 80 degree Celsius to 60 degree celsius in one minute tab indent temperature is is 60 degree in cooling from 60 degree Celsius 50 degree it will take what is the answer options 1) 30 sec
2) 60sec
3) 90sec
4) 50sec

Profile image of Datta Wakde
6 Years agoGrade 12
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2 Answers

Profile image of Vikas TU
6 Years ago
Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To] 
( Tf - Ti)/t = K[ ( 80 + 60)/2 - 60]
( 80-60)/1 = K[ 70 - 60]
20 = K× 10 
K = 20/10 = 2
in second condition, 
initial temperature ( Ti) = 60°C 
Final temperature ( Tf) = 50°C 
Time taken for cooling is t 
A/C Newton's law of cooling 
( 60 - 50)/t = 2 [ (60+50)/2 -60]
10/t = 2* 5 
10/t = 10  
t = 1 min.
Profile image of Khimraj
6 Years ago
According to Rate of cooling
dT/dt  = K((T_initial+T_final)/2 - T_surr) 
( 80-60)/1 = K(70-60)
K = 2
Time taken for cooling
(60 - 50)/t = 2 ((60+50)/2 -60)
t = 60 sec

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